The slope of the common tangent drawn to the | vedclass

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(C) For the circle $C_1: x^2+y^2-4x+12y-216=0$,the center $C_1$ is $(2, -6)$ and radius $r_1 = \sqrt{2^2 + (-6)^2 - (-216)} = \sqrt{4 + 36 + 216} = \s

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$\frac{5}{12}$

Vedclass · Answer

(C) For the circle $C_1: x^2+y^2-4x+12y-216=0$,the center $C_1$ is $(2, -6)$ and radius $r_1 = \sqrt{2^2 + (-6)^2 - (-216)} = \sqrt{4 + 36 + 216} = \sqrt{256} = 16$. For the circle $C_2: x^2+y^2+6x-12y+36=0$,the center $C_2$ is $(-3, 6)$ and radius $r_2 = \sqrt{(-3)^2 + 6^2 - 36} = \sqrt{9 + 36 - 36} = 3$. The distance between centers $d = \sqrt{(2 - (-3))^2 + (-6 - 6)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13$. Since $r_1 - r_2 = 16 - 3 = 13$,which is equal to $d$,the circles touch each other internally. There is only one common tangent at the point of contact. The point of contact $P$ divides the line segment $C_1C_2$ externally in the ratio $r_1 : r_2 = 16 : 3$. $P = \left( \frac{16(-3) - 3(2)}{16 - 3}, \frac{16(6) - 3(-6)}{16 - 3} \right) = \left( \frac{-48 - 6}{13}, \frac{96 + 18}{13} \right) = \left( -\frac{54}{13}, \frac{114}{13} \right)$. The slope of the line $C_1C_2$ is $m_{C_1C_2} = \frac{6 - (-6)}{-3 - 2} = \frac{12}{-5} = -\frac{12}{5}$. The common tangent is perpendicular to the line joining the centers. Therefore,the slope of the common tangent is $m = -\frac{1}{m_{C_1C_2}} = -\frac{1}{-12/5} = \frac{5}{12}$.

The slope of the common tangent drawn to the circles $x^2+y^2-4x+12y-216=0$ and $x^2+y^2+6x-12y+36=0$ is

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